Question

For $x>0,\underset{x\to 0}{\mathrm{Lim}}\left((\tan x{)}^{\frac{1}{x}}+(1+\sin x{)}^x+(\cot x{)}^x\right)$ is equal to

• A. 1
• B. 0
• C. 2
• D. non existent

Answer 1

Answer: (C)

$\underset{x\to 0}{\mathrm{Lim}}(\tan x{)}^{\frac{1}{x}}=\underset{x\to 0}{\mathrm{Lim}}(\tan h{)}^{\frac{1}{h}}=0$
$\underset{x\to 0}{\mathrm{Lim}}(1+\sin x{)}^x=(1+0{)}^0=1$
Let $l=\underset{x\to 0}{\mathrm{Lim}}(\cot x{)}^x({\infty }^0$ form $)$
$\therefore \ln l=\underset{x\to 0}{\mathrm{Lim}}x\ln (\cot x)=\underset{x\to 0}{\mathrm{Lim}}\frac{\ln (\cot x)}{\frac{1}{x}}$
$=\underset{x\to 0}{\mathrm{Lim}}\frac{\frac{-{\mathrm{cosec}}^{2}x}{\cot x}}{\frac{-1}{x^2}}=<br/>\underset{x\to 0}{\mathrm{Lim}}\left(\frac{x^2}{{\sin }^{2}x}\right)\tan x=0$
(Using L'Hospital's Rule)
$\Rightarrow l=1$
Hence $\underset{x\to 0}{\mathrm{Lim}}\left((\tan x{)}^{\frac{1}{x}}+(1+\sin x{)}^x+(\cot x{)}^x\right)=0+1+1=2$